Find points on the ellipse $2x^2-4xy+5y^2=54$ closest to origin using Lagrange multipliers. It's a past paper question.

I let $$f=x^2+y^2$$ and the system of equations I got is $$2x=\lambda (4x-4y)$$$$2y=\lambda (10y-4x)$$ $$2x^2-4xy+5y^2-54=0$$ I eliminated $\lambda $ to get $$\frac yx=\frac{5y-2x}{2x-2y}$$ After which I tirelessly tried to solve for $x$ and $y$ with the constraint (using hand - since it's a past-paper question) and couldn't solve it. Please advice on quicker/efficient ways if any.


Hint: Let $y = xt$, then : $t = \dfrac{5xt - 2x}{2x - 2xt} = \dfrac{5t-2}{2-2t}$. You can solve for $t$, and can go from here.

  • $\begingroup$ This is a very nice trick! Do you think we could always do this given 2 variables, if we know that x is not zero? $\endgroup$
    – NoChance
    Jul 29 '19 at 2:17
  • 1
    $\begingroup$ I think we can😀. $\endgroup$
    – DeepSea
    Jul 29 '19 at 6:28

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